A wheel is free to rotate about its fixed axle. A spring with constant k = 290 N/m is attached to one of its spokes a distance r = 23 cm from the axle, as shown in the figure. (a) Assuming that the wheel is a hoop of mass m = 490 g and radius R = 0.97 m, what is the angular frequency ω of small oscillations of this system? What is ω if (b)r = R and (c)r = 0?
To find the angular frequency ω of small oscillations of the system, we can use the concept of Hooke's Law and the formula for the angular frequency of a simple harmonic oscillator.
a) When the spring is attached to one of the spokes a distance r = 23 cm from the axle, the restoring torque due to the spring force is given by the equation:
τ = -k * r * θ
where τ is the restoring torque, k is the spring constant, r is the distance from the axle to the point of attachment, and θ is the angular displacement.
For small oscillations, we can approximate sin(θ) ≈ θ. Therefore, the equation above simplifies to:
τ = -k * r * θ ≈ -k * r^2 * θ
In rotational motion, the torque τ is also equal to the moment of inertia I multiplied by the angular acceleration α. So we can write:
I * α = -k * r^2 * θ
The moment of inertia of a hoop rotating about its axis is given by:
I = m * R^2
where m is the mass of the hoop and R is its radius.
Substituting this into the equation above, we get:
m * R^2 * α = -k * r^2 * θ
The angular acceleration α can be expressed as:
α = d^2θ/dt^2
where t is time.
So the equation becomes:
m * R^2 * d^2θ/dt^2 = -k * r^2 * θ
This is a second-order linear differential equation. Assuming harmonic solutions of the form:
θ = A * cos(ω * t)
where A is the amplitude of the oscillation and ω is the angular frequency, we can differentiate twice with respect to time to find:
d^2θ/dt^2 = -ω^2 * A * cos(ω * t)
Substituting into the differential equation, we have:
-m * R^2 * ω^2 * A * cos(ω * t) = -k * r^2 * A * cos(ω * t)
Canceling out the common terms, we get:
m * R^2 * ω^2 = k * r^2
Simplifying further, we can solve for ω:
ω = sqrt(k * r^2 / m * R^2)
Plugging in the given values: k = 290 N/m, r = 23 cm = 0.23 m, m = 490 g = 0.49 kg, and R = 0.97 m, we can find ω:
ω = sqrt(290 N/m * (0.23 m)^2 / (0.49 kg * (0.97 m)^2))
Calculating this expression, we find that ω ≈ 9.63 rad/s.
Therefore, the angular frequency ω of small oscillations of this system when r = 23 cm is approximately 9.63 rad/s.
b) If r = R, the equation for the angular frequency becomes:
ω = sqrt(k * R^2 / m * R^2)
Simplifying further, we have:
ω = sqrt(k / m)
Plugging in the given values of k = 290 N/m and m = 0.49 kg, we can calculate ω:
ω = sqrt(290 N/m / 0.49 kg)
Calculating this expression, we find that ω ≈ 17.28 rad/s.
Therefore, the angular frequency ω of small oscillations of this system when r = R is approximately 17.28 rad/s.
c) If r = 0, the equation for the angular frequency becomes:
ω = sqrt(k * 0 / m * R^2)
Since r = 0, there is no spring force acting on the system, and the wheel will not experience any restoring torque. As a result, there will be no oscillations and the angular frequency ω will be zero.
Therefore, the angular frequency ω of small oscillations of this system when r = 0 is zero.